Algebra - Surds & Indices - Previous Year CAT/MBA Questions

The best way to prepare for Algebra - Surds & Indices is by going through the previous year Algebra - Surds & Indices OMET questions. Here we bring you all previous year Algebra - Surds & Indices OMET questions along with detailed solutions.

It would be best if you clear your concepts before you practice previous year Algebra - Surds & Indices OMET questions.

1. XAT 2019 QADI | Algebra - Surds & Indices OMET Question

If $\sqrt[3]{7^{a} \times (35)^{(b + 1)} \times (20)^{(c + 2)}}$ is a whole number then which one of the statements below is consistent with it?

(a)
a = 3, b = 2, c = 1
(b)
a = 3, b = 1, c = 1
(c)
a = 2, b = 1, c = 2
(d)
a = 2, b = 1, c = 1
(e)
a = 1, b = 2, c = 2

2. IIFT 2019 QA | Algebra - Surds & Indices OMET Question

If x = 8 - $\sqrt{32}$ and y = 2 + √2, then ${(x + \frac{1}{y})}^{2}$ is given by:

(a)
16x²/25
(b)
64x²/81
(c)
25y²/16
(d)
81x²/64

Answer: Option D

Explanation :

x = 8 − √32 = 8 − 4√2 = 4(2 − √2)

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}}$

On rationalising, we get

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ = $\frac{2 - \sqrt{2}}{2}$ = $\frac{x}{8}$

∴ x + 1/y = x + x/8 = 9x/8

⇒ ${(x + \frac{1}{y})}^{2}$ = ${(x + \frac{x}{8})}^{2}$ = ${(\frac{9 x}{8})}^{2}$ = $\frac{81 x^{2}}{64}$

Hence option (d).

3. IIFT 2016 QA | Algebra - Surds & Indices OMET Question

The highest number amongst $\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[4]{4}$

(a)
$\sqrt{2}$
(b)
$\sqrt[3]{3}$
(c)
$\sqrt[4]{4}$
(d)
None of the above

4. IIFT 2015 QA | Algebra - Surds & Indices OMET Question

The value of x for which the equation $\sqrt{4 x - 9} + \sqrt{4 x + 9} = 5 + \sqrt{7}$ will be satisfied is:

(a)
1
(b)
2
(c)
3
(d)
4

5. IIFT 2015 QA | Algebra - Surds & Indices OMET Question

The simplest value of the expression ${\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\}}^{\frac{1}{p}}$ is:

(a)
4
(b)
8
(c)
4^p
(d)
8^p

Answer: Option B

Explanation :

$\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\} = \{\frac{4^{p} \times 4^{\frac{1}{4}} \times \sqrt{2} \times \sqrt{2^{p}}}{2 \times \sqrt{2^{- p}}}\} = \{\frac{2^{2 p} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2^{p}}}{2 \times \sqrt{2^{- p}}}\} = 2^{2 p} \times \sqrt{2^{p}} \times \sqrt{2^{p}} = 2^{3 p}$

${\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\}}^{\frac{1}{p}} = {(2^{3 p})}^{\frac{1}{p}} = 8$

Hence, option (b).

Algebra - Surds & Indices - Previous Year CAT/MBA Questions

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